1
$\begingroup$

I am trying to find the eigenvalues and eigenfunctions of the integral operator $Ku=\int_0^\pi k(x,y)u(y)dy$ with the following kernel: $k( x,y) = \sum\limits_{n=1}^\infty \frac{1}{n^2} \sin\big((n+1)x\big)\sin(ny)$.

Using the DCT we can exchange the sum and the integral to get: $$\sum\limits_{n=1}^\infty \frac{1}{n^2} \sin\big((n+1)x\big) \int_0^{\pi}\sin(ny)u(y)dy=\lambda u(y)$$

Now the LHS looks like a Fourier series of a function but I cannot guess which one. Any hints?

P.S. There is exactly the same question here Find eigenfunctions of the integral operator with kernel $\sum\limits_{n=1}^\infty \frac{1}{n^2} \sin((n+1)x)\sin(ny)$, with no answer and I followed the given hint there by replacing $u$ with its Fourier series but the integral terms argued there to be zero are definitely wrong.

$\endgroup$
1
$\begingroup$

The functions $\{ \sin(nx) \}_{n=1}^{\infty}$ form a complete orthogonal basis for $L^2(0,\pi)$ because they are the eigenfunction solutions of $$ -y'' = \lambda y,\;\;\; y(0)=y(\pi)=0. $$ The normalization constants are $$ \int_{0}^{\pi}\sin^2(nx)dx=\frac{1}{2}\int_{0}^{2\pi}\sin^2(nx)dx = \frac{1}{4}\int_{0}^{2\pi}\sin^2(nx)+\cos^2(nx)dx = \frac{\pi}{2}. $$ Any $f\in L^2(0,\pi)$ can be written uniquely as $\sum_{n=1}^{\infty}f_n\sin(nx)$ where $\{ f_n \}_{n=1}^{\infty} \in \ell^2$. In fact, $$ f_n = \frac{2}{\pi}\int_{0}^{\pi}f(x)\sin(nx)dx,\;\; n=1,2,3,\cdots. $$ So your problem can be reduced to a problem on $\ell^2(\mathbb{N})$, where $u$ is represented by $\{ u_n \}_{n=1}^{\infty}$. The eigenfunction problem becomes a coefficient problem in $\ell^2$: $$ Ku=\lambda u \\ \int_{0}^{\pi}u(y)k(x,y)dy=\lambda u(x) \\ \int_{0}^{\pi}u(y)\sum_{n=1}^{\infty}\frac{1}{n^2}\sin((n+1)x)\sin(ny)dy=\lambda u(x) $$ This gives coefficient equations after rewriting as \begin{align} \sum_{n=1}^{\infty}\frac{1}{n^2}\sin((n+1)x)\frac{\pi}{2}u_n&=\sum_{n=1}^{\infty}\lambda u_n\sin(nx) \\ \sum_{n=2}^{\infty}\frac{1}{(n-1)^2}\sin(nx)\frac{\pi}{2}u_{n-1} & = \sum_{n=1}^\infty \lambda u_n \sin(nx). \end{align} There is no $\sin(x)$ term on the left, which forces $\lambda u_1=0$. The general $u_n$ must satisfy $$ \frac{\pi}{2}\frac{1}{(n-1)^2}u_{n-1}=\lambda u_n,\;\; n \ge 2. $$ If $\lambda \ne 0$, then $u_1 =0$ and every $u_n=0$ for $n > 1$ by the above. If $\lambda = 0$, then every $u_n=0$ by the above. So there are no eigenvalues of this operataor. That means that the operator $K$ is quasinilpotent with spectrum $\{0\}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.